Functional Equations Questions in English

(D) Given the functional equation $f(xy) = f(x) + f(y)$.
We know $f(8) = f(2 \cdot 2 \cdot 2) = f(2) + f(2) + f(2) = 3f(2)$.
Given $f(8) = 15$,so $3f(2) = 15$,which implies $f(2) = 5$.
Now,we need to find $f(48)$.
We can write $48 = 12 \cdot 4 = 12 \cdot 2 \cdot 2$.
Using the property $f(xy) = f(x) + f(y)$,we get:
$f(48) = f(12 \cdot 2 \cdot 2) = f(12) + f(2) + f(2)$.
Substituting the known values $f(12) = 24$ and $f(2) = 5$:
$f(48) = 24 + 5 + 5 = 34$.
Thus,the correct option is $D$.

(C) Given equation is $f(x) + (x + \frac{1}{2}) f(1 - x) = 1$ $(i)$
Replacing $x$ with $1 - x$ in $(i)$,we get:
$f(1 - x) + (1 - x + \frac{1}{2}) f(x) = 1$
$f(1 - x) + (\frac{3}{2} - x) f(x) = 1$ $(ii)$
From $(ii)$,$f(1 - x) = 1 - (\frac{3}{2} - x) f(x)$.
Substitute this into $(i)$:
$f(x) + (x + \frac{1}{2}) [1 - (\frac{3}{2} - x) f(x)] = 1$
$f(x) + (x + \frac{1}{2}) - (x + \frac{1}{2})(\frac{3}{2} - x) f(x) = 1$
$f(x) [1 - (\frac{3}{2}x - x^2 + \frac{3}{4} - \frac{1}{2}x)] = 1 - (x + \frac{1}{2})$
$f(x) [1 - (-x^2 + x + \frac{3}{4})] = \frac{1}{2} - x$
$f(x) [x^2 - x + \frac{1}{4}] = \frac{1}{2} - x$
$f(x) (x - \frac{1}{2})^2 = - (x - \frac{1}{2})$
For $x \neq \frac{1}{2}$,$f(x) = \frac{-(x - \frac{1}{2})}{(x - \frac{1}{2})^2} = \frac{-1}{x - \frac{1}{2}} = \frac{2}{1 - 2x}$.
Now,$f(0) = \frac{2}{1 - 0} = 2$ and $f(1) = \frac{2}{1 - 2} = -2$.
Therefore,$2 f(0) + 3 f(1) = 2(2) + 3(-2) = 4 - 6 = -2$.

(B) Given the functional equation $f(x) + f(2x) = 0$ for all $x \in R$.
Substituting $x = 0$,we get $f(0) + f(0) = 0$,which implies $2f(0) = 0$,so $f(0) = 0$.
From the given equation,$f(x) = -f(x/2)$.
By iterating this relation,we get $f(x) = -f(x/2) = f(x/4) = -f(x/8) = \dots = (-1)^n f(x/2^n)$.
Since $f$ is continuous at $x = 0$,we have $\lim_{n \to \infty} f(x/2^n) = f(\lim_{n \to \infty} x/2^n) = f(0) = 0$.
Thus,$f(x) = \lim_{n \to \infty} (-1)^n f(x/2^n) = 0$.
Therefore,the only continuous function satisfying the condition is the zero function $f(x) = 0$.
Hence,there is only $1$ such function.

(D) Given the relation $f(n) + \frac{1}{n} f(n+1) = 1$,we can write $f(n+1) = n(1 - f(n))$.
For $n=3$: $f(4) = 3(1 - f(3))$. Since $f(4) \in \mathbb{Z}$ and $|f(4)| \leq 8$,$f(4)$ must be a multiple of $3$ such that $|f(4)| \leq 8$. Possible values for $f(4)$ are $\{-6, -3, 0, 3, 6\}$.
For $n=2$: $f(3) = 2(1 - f(2))$. Thus $f(3)$ must be an even integer such that $|f(3)| \leq 8$. Possible values for $f(3)$ are $\{-8, -6, -4, -2, 0, 2, 4, 6, 8\}$.
For $n=1$: $f(2) = 1(1 - f(1)) = 1 - f(1)$. Thus $f(2)$ can be any integer such that $|f(2)| \leq 8$.
We check the constraints:
$1$. $f(4) = 3(1 - f(3)) \implies f(3) = 1 - \frac{f(4)}{3}$. For $f(3)$ to be an even integer,$\frac{f(4)}{3}$ must be an odd integer. Thus $f(4) \in \{-3, 3\}$.
$2$. If $f(4) = -3$,then $f(3) = 1 - (-1) = 2$. Then $f(2) = 1 - \frac{f(3)}{2} = 1 - 1 = 0$. Then $f(1) = 1 - f(2) = 1 - 0 = 1$. All values are in the range $[-8, 8]$.
$3$. If $f(4) = 3$,then $f(3) = 1 - 1 = 0$. Then $f(2) = 1 - \frac{f(3)}{2} = 1 - 0 = 1$. Then $f(1) = 1 - f(2) = 1 - 1 = 0$. All values are in the range $[-8, 8]$.
Thus,there are $2$ such functions.

(C) Given $f(x + y) = f(x) + f(y)$,this is Cauchy's functional equation on $\mathbb{N}$,which implies $f(n) = cn$ for some constant $c$.
Since $f(1) = \frac{1}{5}$,we have $c(1) = \frac{1}{5}$,so $f(n) = \frac{n}{5}$.
Now,substitute $f(n) = \frac{n}{5}$ into the summation:
$\sum_{n=1}^m \frac{n/5}{n(n+1)(n+2)} = \frac{1}{5} \sum_{n=1}^m \frac{1}{(n+1)(n+2)}$.
Using partial fractions,$\frac{1}{(n+1)(n+2)} = \frac{1}{n+1} - \frac{1}{n+2}$.
So,the sum becomes $\frac{1}{5} \sum_{n=1}^m \left( \frac{1}{n+1} - \frac{1}{n+2} \right)$.
This is a telescoping sum:
$\frac{1}{5} \left( (\frac{1}{2} - \frac{1}{3}) + (\frac{1}{3} - \frac{1}{4}) + \dots + (\frac{1}{m+1} - \frac{1}{m+2}) \right) = \frac{1}{5} \left( \frac{1}{2} - \frac{1}{m+2} \right)$.
Given the sum is $\frac{1}{12}$,we have $\frac{1}{5} \left( \frac{m+2-2}{2(m+2)} \right) = \frac{1}{12} \implies \frac{m}{10(m+2)} = \frac{1}{12}$.
$12m = 10m + 20 \implies 2m = 20 \implies m = 10$.

(C) Given the functional equation $f(x + y) = f(x) + f(y) - 1$.
To find $f(0)$,put $x = 0$ and $y = 0$:
$f(0 + 0) = f(0) + f(0) - 1 \Rightarrow f(0) = 2f(0) - 1 \Rightarrow f(0) = 1$.
Now,use the definition of the derivative:
$f'(x) = \lim_{h \rightarrow 0} \frac{f(x + h) - f(x)}{h}$.
Using the given relation $f(x + h) = f(x) + f(h) - 1$:
$f'(x) = \lim_{h \rightarrow 0} \frac{f(x) + f(h) - 1 - f(x)}{h} = \lim_{h \rightarrow 0} \frac{f(h) - 1}{h}$.
Since $f(0) = 1$,we can write $1 = f(0)$:
$f'(x) = \lim_{h \rightarrow 0} \frac{f(h) - f(0)}{h} = f'(0)$.
Given $f'(0) = 2$,so $f'(x) = 2$.
Integrating both sides with respect to $x$:
$f(x) = 2x + C$.
Using $f(0) = 1$:
$1 = 2(0) + C \Rightarrow C = 1$.
Thus,$f(x) = 2x + 1$.
Now,calculate $f(-2)$:
$f(-2) = 2(-2) + 1 = -4 + 1 = -3$.
Therefore,$|f(-2)| = |-3| = 3$.

(C) Given the functional equation $5f(x + y) = f(x) \cdot f(y)$.
Setting $x = 0$ and $y = 0$,we get $5f(0) = f(0)^2$. Since the codomain is $(0, \infty)$,$f(0) \neq 0$,so $f(0) = 5$.
Setting $y = 1$,we have $5f(x + 1) = f(x) \cdot f(1)$,which implies $\frac{f(x + 1)}{f(x)} = \frac{f(1)}{5}$.
This shows that $f(n)$ is a geometric progression with common ratio $r = \frac{f(1)}{5}$.
We know $f(3) = f(0) \cdot r^3 = 5 \cdot r^3 = 320$.
Thus,$r^3 = 64$,which gives $r = 4$.
Therefore,$f(n) = f(0) \cdot r^n = 5 \cdot 4^n$.
The sum is $\sum_{n=0}^5 f(n) = \sum_{n=0}^5 5 \cdot 4^n = 5(1 + 4 + 4^2 + 4^3 + 4^4 + 4^5)$.
Using the geometric series sum formula $S_n = a\frac{r^n - 1}{r - 1}$,we get $5 \cdot \frac{4^6 - 1}{4 - 1} = 5 \cdot \frac{4096 - 1}{3} = 5 \cdot \frac{4095}{3} = 5 \cdot 1365 = 6825$.

(B) Given the functional equation: $f(x) + f\left(\frac{1}{1-x}\right) = 1 + x$ $(1)$
Substitute $x = 2$ in $(1)$:
$f(2) + f\left(\frac{1}{1-2}\right) = 1 + 2$
$f(2) + f(-1) = 3$ $(2)$
Substitute $x = -1$ in $(1)$:
$f(-1) + f\left(\frac{1}{1-(-1)}\right) = 1 + (-1)$
$f(-1) + f\left(\frac{1}{2}\right) = 0$ $(3)$
Substitute $x = \frac{1}{2}$ in $(1)$:
$f\left(\frac{1}{2}\right) + f\left(\frac{1}{1-1/2}\right) = 1 + \frac{1}{2}$
$f\left(\frac{1}{2}\right) + f(2) = \frac{3}{2}$ $(4)$
Now,perform $(2) + (4) - (3)$:
$(f(2) + f(-1)) + (f\left(\frac{1}{2}\right) + f(2)) - (f(-1) + f\left(\frac{1}{2}\right)) = 3 + \frac{3}{2} - 0$
$2f(2) = \frac{9}{2}$
$f(2) = \frac{9}{4}$

(A) Given the functional equation $f\left(\frac{x}{y}\right)=\frac{f(x)}{f(y)}$.
Setting $x=y=1$,we get $f(1) = \frac{f(1)}{f(1)} = 1$.
Now,differentiate the given equation partially with respect to $x$:
$f^{\prime}\left(\frac{x}{y}\right) \cdot \frac{1}{y} = \frac{f^{\prime}(x)}{f(y)}$.
Substitute $y=x$ into the equation:
$f^{\prime}(1) \cdot \frac{1}{x} = \frac{f^{\prime}(x)}{f(x)}$.
Since $f^{\prime}(1) = 2024$,we have:
$2024 \cdot \frac{1}{x} = \frac{f^{\prime}(x)}{f(x)}$.
Rearranging the terms,we get:
$x f^{\prime}(x) = 2024 f(x)$,
which implies $x f^{\prime}(x) - 2024 f(x) = 0$.

(A) Given the functional equation $f(m+n) = f(m) + f(n)$ for all $m, n \in \mathbb{N}$.
This is Cauchy's functional equation on natural numbers,which implies $f(x) = cx$.
Given $f(1) = 1$,we have $c(1) = 1$,so $c = 1$.
Thus,$f(x) = x$.
Now,we need to solve the inequality $\sum_{k=1}^{2022} f(\lambda+k) \leq (2022)^2$.
Substituting $f(x) = x$,we get $\sum_{k=1}^{2022} (\lambda+k) \leq (2022)^2$.
Expanding the summation: $2022\lambda + \sum_{k=1}^{2022} k \leq (2022)^2$.
Using the sum formula $\sum_{k=1}^{n} k = \frac{n(n+1)}{2}$,we have $2022\lambda + \frac{2022 \times 2023}{2} \leq (2022)^2$.
Dividing by $2022$: $\lambda + \frac{2023}{2} \leq 2022$.
$\lambda \leq 2022 - 1011.5$.
$\lambda \leq 1010.5$.
Since $\lambda$ must be a natural number,the largest value is $\lambda = 1010$.

(B, C) Given the functional equation $f(x+y)=f(x)+f(y)$,which is Cauchy's functional equation.
Setting $x=0, y=0$,we get $f(0)=f(0)+f(0)$,which implies $f(0)=0$.
Since $f(x)$ is differentiable at $x=0$,we have $f^{\prime}(0) = \lim_{h \rightarrow 0} \frac{f(0+h)-f(0)}{h} = \lim_{h \rightarrow 0} \frac{f(h)}{h} = k$ (where $k$ is a constant).
Now,for any $x \in R$,the derivative $f^{\prime}(x)$ is given by:
$f^{\prime}(x) = \lim_{h \rightarrow 0} \frac{f(x+h)-f(x)}{h} = \lim_{h \rightarrow 0} \frac{f(x)+f(h)-f(x)}{h} = \lim_{h \rightarrow 0} \frac{f(h)}{h} = f^{\prime}(0) = k$.
Since $f^{\prime}(x) = k$ for all $x \in R$,$f^{\prime}(x)$ is a constant function.
Integrating $f^{\prime}(x) = k$,we get $f(x) = kx + C$. Since $f(0)=0$,we have $C=0$,so $f(x) = kx$.
$A$ linear function $f(x) = kx$ is continuous and differentiable for all $x \in R$.
Thus,statements $(B)$ and $(C)$ are true.

(C) Given $f(x+y)=f(x)+f(y)$. This is Cauchy's functional equation,which implies $f(x)=ax$ for all $x \in Q$.
Given $f\left(\frac{-3}{5}\right)=12$,we have $a\left(\frac{-3}{5}\right)=12$,so $a=-20$. Thus,$f(x)=-20x$.
Then $f\left(\frac{1}{4}\right)=-20\left(\frac{1}{4}\right)=-5$.
Given $g(x+y)=g(x)g(y)$,this implies $g(x)=b^x$ for some $b>0$.
Given $g\left(\frac{-1}{3}\right)=2$,we have $b^{-1/3}=2$,which means $b^{-1}=2^3=8$,so $b=\frac{1}{8}$.
Thus,$g(x)=\left(\frac{1}{8}\right)^x=8^{-x}=2^{-3x}$.
Then $g(-2)=2^{-3(-2)}=2^6=64$.
Also,$g(0)=b^0=1$.
Substituting these values into the expression: $\left(f\left(\frac{1}{4}\right)+g(-2)-8\right)g(0) = (-5+64-8)(1) = 51$.

(NONE) Given $f(x) = (3a+2)x^2 + \left(\frac{a+2}{a-1}\right)x + b$.
Putting $x=y=0$ in $f(x+y) = f(x) + f(y) + 1 - \frac{2}{7}xy$,we get $f(0) = 2f(0) + 1$,which implies $f(0) = -1$.
Since $f(0) = b$,we have $b = -1$.
Now,$f(x+y) = f(x) + f(y) + 1 - \frac{2}{7}xy$.
Comparing the coefficients of $xy$ in the expansion of $f(x+y) = A(x+y)^2 + B(x+y) + C$,we have $f(x+y) = A(x^2+2xy+y^2) + B(x+y) + C = Ax^2 + Ay^2 + 2Axy + Bx + By + C$.
Given $f(x+y) = f(x) + f(y) + 1 - \frac{2}{7}xy$,we equate $2A = -\frac{2}{7}$,so $A = -\frac{1}{7}$.
Since $A = 3a+2$,we have $3a+2 = -\frac{1}{7} \Rightarrow 3a = -\frac{1}{7} - 2 = -\frac{15}{7} \Rightarrow a = -\frac{5}{7}$.
Now,$B = \frac{a+2}{a-1} = \frac{-5/7 + 2}{-5/7 - 1} = \frac{9/7}{-12/7} = -\frac{3}{4}$.
Thus,$f(x) = -\frac{1}{7}x^2 - \frac{3}{4}x - 1$.
We need to calculate $28 \sum_{i=1}^3 |f(i)|$.
$f(1) = -\frac{1}{7} - \frac{3}{4} - 1 = \frac{-4-21-28}{28} = -\frac{53}{28}$.
$f(2) = -\frac{4}{7} - \frac{6}{4} - 1 = -\frac{4}{7} - \frac{3}{2} - 1 = \frac{-8-21-14}{14} = -\frac{43}{14} = -\frac{86}{28}$.
$f(3) = -\frac{9}{7} - \frac{9}{4} - 1 = \frac{-36-63-28}{28} = -\frac{127}{28}$.
$28 \sum_{i=1}^3 |f(i)| = 28 \left( \frac{53}{28} + \frac{86}{28} + \frac{127}{28} \right) = 53 + 86 + 127 = 266$.

(B) Given $f(x)f(\frac{1}{x}) = f(x) + f(\frac{1}{x})$.
Let $f(x) = ax^2 + bx + c$. Since $f(x)$ is a polynomial of degree $2$,$a \neq 0$.
Substituting $f(x)$ into the equation: $(ax^2 + bx + c)(a(\frac{1}{x})^2 + b(\frac{1}{x}) + c) = ax^2 + bx + c + a(\frac{1}{x})^2 + b(\frac{1}{x}) + c$.
Comparing coefficients,we find $c = 1$ and $a = -1$ (since the range is $(-\infty, 1)$).
Thus,$f(x) = 1 - x^2$.
Given $f(K) = -2K$,we have $1 - K^2 = -2K$,which simplifies to $K^2 - 2K - 1 = 0$.
Let the roots be $K_1$ and $K_2$. The sum of squares is $K_1^2 + K_2^2 = (K_1 + K_2)^2 - 2K_1K_2$.
Using Vieta's formulas,$K_1 + K_2 = 2$ and $K_1K_2 = -1$.
Therefore,$K_1^2 + K_2^2 = (2)^2 - 2(-1) = 4 + 2 = 6$.

(B) Given the equation: $(\sin x \cos y)(f(2x+2y) - f(2x-2y)) = (\cos x \sin y)(f(2x+2y) + f(2x-2y))$.
Rearranging the terms,we get: $f(2x+2y)(\sin x \cos y - \cos x \sin y) = f(2x-2y)(\sin x \cos y + \cos x \sin y)$.
Using the trigonometric identities $\sin(A-B) = \sin A \cos B - \cos A \sin B$ and $\sin(A+B) = \sin A \cos B + \cos A \sin B$,we have: $f(2x+2y)\sin(x-y) = f(2x-2y)\sin(x+y)$.
This can be written as: $\frac{f(2x+2y)}{\sin(x+y)} = \frac{f(2x-2y)}{\sin(x-y)}$.
Let $2x+2y = m$ and $2x-2y = n$. Then $x+y = m/2$ and $x-y = n/2$.
The equation becomes: $\frac{f(m)}{\sin(m/2)} = \frac{f(n)}{\sin(n/2)} = K$ (a constant).
Thus,$f(x) = K \sin(x/2)$.
Differentiating with respect to $x$,we get $f'(x) = \frac{K}{2} \cos(x/2)$.
Given $f'(0) = 1/2$,we have $\frac{K}{2} \cos(0) = 1/2$,which implies $K = 1$.
So,$f(x) = \sin(x/2)$.
Then $f'(x) = \frac{1}{2} \cos(x/2)$ and $f''(x) = -\frac{1}{4} \sin(x/2)$.
We need to find $24f''\left(\frac{5\pi}{3}\right) = 24 \left(-\frac{1}{4} \sin\left(\frac{5\pi}{6}\right)\right)$.
Since $\sin(5\pi/6) = 1/2$,the value is $24 \left(-\frac{1}{4} \times \frac{1}{2}\right) = 24 \left(-\frac{1}{8}\right) = -3$.

(A) Given the equation: $f(x) + 3f\left(\frac{24}{x}\right) = 4x$
Step $1$: Substitute $x = 3$ into the equation:
$f(3) + 3f\left(\frac{24}{3}\right) = 4(3)$
$f(3) + 3f(8) = 12$ --- (Equation $1$)
Step $2$: Substitute $x = 8$ into the equation:
$f(8) + 3f\left(\frac{24}{8}\right) = 4(8)$
$f(8) + 3f(3) = 32$ --- (Equation $2$)
Step $3$: Add Equation $1$ and Equation $2$:
$(f(3) + 3f(8)) + (f(8) + 3f(3)) = 12 + 32$
$4f(3) + 4f(8) = 44$
$4(f(3) + f(8)) = 44$
Step $4$: Divide by $4$:
$f(3) + f(8) = 11$

(C) Given $h(x) = f(x+1) - g(x+2)$.
Expanding the terms involving $x^3$ in $h(x)$:
$f(x+1) = a_1 + 10(x+1) + a_2(x+1)^2 + a_3(x+1)^3 + (x+1)^4$
The coefficient of $x^3$ in $f(x+1)$ is $a_3 + 4(1) = a_3 + 4$.
$g(x+2) = b_1 + 3(x+2) + b_2(x+2)^2 + b_3(x+2)^3 + (x+2)^4$
The coefficient of $x^3$ in $g(x+2)$ is $b_3 + 4(2) = b_3 + 8$.
Thus,the coefficient of $x^3$ in $h(x)$ is $(a_3 + 4) - (b_3 + 8) = a_3 - b_3 - 4$.
Since $f(x) - g(x) \neq 0$ for all $x \in R$,the expression $f(x) - g(x) = (a_3 - b_3)x^3 + (a_2 - b_2)x^2 + 7x + (a_1 - b_1)$ must not have any real roots.
For a cubic polynomial to have no real roots,the coefficient of $x^3$ must be $0$ (otherwise,it would be a cubic equation which always has at least one real root).
Therefore,$a_3 - b_3 = 0$.
Substituting this into the coefficient of $x^3$ in $h(x)$,we get $0 - 4 = -4$.

(A) Given the equation: $3 f(x) - f\left(\frac{1}{x}\right) = 8 \log_2 x^3 = 24 \log_2 x$ $(i)$
Replace $x$ with $\frac{1}{x}$: $3 f\left(\frac{1}{x}\right) - f(x) = 24 \log_2 \left(\frac{1}{x}\right) = -24 \log_2 x$ $(ii)$
Multiply $(i)$ by $3$: $9 f(x) - 3 f\left(\frac{1}{x}\right) = 72 \log_2 x$ $(iii)$
Add $(ii)$ and $(iii)$: $(9 f(x) - f(x)) + (3 f\left(\frac{1}{x}\right) - 3 f\left(\frac{1}{x}\right)) = 72 \log_2 x - 24 \log_2 x$
$8 f(x) = 48 \log_2 x \Rightarrow f(x) = 6 \log_2 x$
Now,calculate the values:
$f(2) = 6 \log_2 2 = 6(1) = 6$
$f(4) = 6 \log_2 4 = 6(2) = 12$
$f(8) = 6 \log_2 8 = 6(3) = 18$
Since $12 - 6 = 6$ and $18 - 12 = 6$,the common difference is constant.
Therefore,$f(2), f(4), f(8)$ are in $A$.$P$.

(D) We are given the functional equation $f(x+y)=f(x) \cdot f(y)$.
By the definition of the derivative,$f^{\prime}(3)=\lim _{h \rightarrow 0} \frac{f(3+h)-f(3)}{h}$.
Using the given functional equation,$f(3+h)=f(3) \cdot f(h)$.
Substituting this into the limit expression:
$f^{\prime}(3)=\lim _{h \rightarrow 0} \frac{f(3) \cdot f(h)-f(3)}{h} = f(3) \lim _{h \rightarrow 0} \frac{f(h)-1}{h}$.
Since $f(0+0)=f(0) \cdot f(0)$,we have $f(0)=f(0)^2$,which implies $f(0)=1$ (assuming $f(x)$ is not identically zero).
Thus,$f^{\prime}(0)=\lim _{h \rightarrow 0} \frac{f(h)-f(0)}{h} = \lim _{h \rightarrow 0} \frac{f(h)-1}{h} = 11$.
Substituting $f(3)=3$ and $\lim _{h \rightarrow 0} \frac{f(h)-1}{h} = 11$ into the expression for $f^{\prime}(3)$:
$f^{\prime}(3) = 3 \times 11 = 33$.

(A) Given $f(x) = x^{2} - 3x + 4$.
First,we find $f(2x + 1)$:
$f(2x + 1) = (2x + 1)^{2} - 3(2x + 1) + 4$
$= (4x^{2} + 4x + 1) - 6x - 3 + 4$
$= 4x^{2} - 2x + 2$.
Since $f(x) = f(2x + 1)$,we equate the two expressions:
$x^{2} - 3x + 4 = 4x^{2} - 2x + 2$
$3x^{2} + x - 2 = 0$
Factoring the quadratic equation:
$3x^{2} + 3x - 2x - 2 = 0$
$3x(x + 1) - 2(x + 1) = 0$
$(x + 1)(3x - 2) = 0$
Thus,$x = -1$ or $x = \frac{2}{3}$.

(B) Given that $f(10-x) = 3x^2 + 4x - 5$ and $f(x) = px^2 + qx + r$.
We need to find $p+q+r$.
Note that $p+q+r = f(1)$.
To find $f(1)$,we set $10-x = 1$,which gives $x = 9$.
Substituting $x = 9$ into the given equation:
$f(10-9) = 3(9)^2 + 4(9) - 5$
$f(1) = 3(81) + 36 - 5$
$f(1) = 243 + 36 - 5$
$f(1) = 279 - 5 = 274$.
Therefore,$p+q+r = 274$.

(B) Given the Cauchy functional equation $f(x+y)=f(x)+f(y)$,the solution is of the form $f(x)=cx$.
Given $f(1)=7$,we have $c(1)=7$,so $c=7$.
Thus,$f(x)=7x$.
We need to calculate $\sum_{t=1}^{39} f(t) = \sum_{t=1}^{39} 7t$.
This is equal to $7 \times \sum_{t=1}^{39} t$.
Using the sum formula $\sum_{k=1}^{n} k = \frac{n(n+1)}{2}$,we get:
$\sum_{t=1}^{39} t = \frac{39 \times 40}{2} = 39 \times 20 = 780$.
Therefore,$\sum_{t=1}^{39} f(t) = 7 \times 780 = 5460$.

(C) Given the relation $f(x f(y)) = f(x y) + x$.
Since $f(x) - x = \lambda$,we have $f(x) = x + \lambda$.
Substituting this into the given functional equation:
$f(x(y + \lambda)) = (xy + \lambda) + x$
$x(y + \lambda) + \lambda = xy + \lambda + x$
$xy + x\lambda + \lambda = xy + \lambda + x$
Comparing the terms,we get $x\lambda = x$,which implies $\lambda = 1$.
Thus,$f(x) = x + 1$.
Now,we evaluate the limit:
$\lim _{x \rightarrow 0} \frac{(x + 1)^{\frac{1}{3}} - 1}{(x + 1)^{\frac{1}{2}} - 1}$
Using the standard limit formula $\lim _{u \rightarrow 1} \frac{u^n - 1}{u - 1} = n$:
$\lim _{x}$ ${\rightarrow 0} \frac{(1 + x)^{\frac{1}{3}} - 1}{(1 + x) - 1} \cdot \frac{(1 + x) - 1}{(1 + x)^{\frac{1}{2}} - 1} = \frac{1/3}{1/2} = \frac{2}{3}$.

(A) Given that: $f(x) = \frac{2x+1}{3}$ ...$(i)$
Also,$f(\alpha) = \frac{1}{\alpha}$.
Substituting $\alpha$ in the function definition:
$\frac{2\alpha+1}{3} = \frac{1}{\alpha}$
$\Rightarrow 2\alpha^2 + \alpha = 3$
$\Rightarrow 2\alpha^2 + \alpha - 3 = 0$
Factoring the quadratic equation:
$2\alpha^2 + 3\alpha - 2\alpha - 3 = 0$
$\Rightarrow \alpha(2\alpha + 3) - 1(2\alpha + 3) = 0$
$\Rightarrow (\alpha - 1)(2\alpha + 3) = 0$
Thus,the possible values for $\alpha$ are $\alpha = 1$ and $\alpha = -\frac{3}{2}$.
The sum of all possible values of $\alpha$ is $1 + (-\frac{3}{2}) = 1 - \frac{3}{2} = -\frac{1}{2}$.

(D) Given $f(g(x+y)) = f(g(x)) + f(g(y))$ ... $(i)$
Let $h(x) = f(g(x))$. Then the equation becomes $h(x+y) = h(x) + h(y)$,which is Cauchy's functional equation.
The solution to this is $h(x) = cx$ for some constant $c$.
Given $g(1) = 2$ and $f(2) = 1$,we have $h(1) = f(g(1)) = f(2) = 1$.
Since $h(1) = c(1) = 1$,we get $c = 1$.
Thus,$h(x) = f(g(x)) = x$.
Since $f(g(x)) = x$,$f$ and $g$ are inverse functions of each other.
Therefore,$g(f(x)) = x$ for all $x \in R$.
The function $g(f(x)) = x$ is a polynomial function,which is continuous everywhere on the set of real numbers $R$.
Hence,the set of points where $g(f(x))$ is discontinuous is the empty set,denoted by $\phi$.

(B) Given the functional equation $f(x+y) = f(x) + f(y)$ for all $x, y \in Z$.
This is Cauchy's functional equation on the set of integers $Z$.
The general solution is $f(x) = kx$ for some constant $k \in Z$.
For $f$ to be a bijection,it must be both injective and surjective.
If $f(x) = kx$,then $f$ is injective if $kx_1 = kx_2 \implies x_1 = x_2$,which holds if $k \neq 0$.
For $f$ to be surjective,for any $y \in Z$,there must exist $x \in Z$ such that $f(x) = y$,i.e.,$kx = y$.
This implies $x = y/k$. For $x$ to be an integer for every $y \in Z$,$k$ must be a divisor of $1$.
Thus,$k = 1$ or $k = -1$.
Therefore,the possible functions are $f(x) = x$ and $f(x) = -x$.
There are exactly $2$ such bijections.

(B) Given,$f(x) = x^{9} - 11 x^{8} - 2 x^{7} + 22 x^{6} + x^{4} - 12 x^{3} + 11 x^{2} + x - 3$.
To find $f(11)$,substitute $x = 11$ into the expression:
$f(11) = 11^{9} - 11(11)^{8} - 2(11)^{7} + 22(11)^{6} + 11^{4} - 12(11)^{3} + 11(11)^{2} + 11 - 3$.
Observe the terms:
$11^{9} - 11(11)^{8} = 11^{9} - 11^{9} = 0$.
$-2(11)^{7} + 22(11)^{6} = -2(11)^{7} + 2(11)(11)^{6} = -2(11)^{7} + 2(11)^{7} = 0$.
$11^{4} - 12(11)^{3} + 11(11)^{2} = 11^{4} - 12(11)^{3} + 11^{3} = 11^{4} - 11(11)^{3} = 11^{4} - 11^{4} = 0$.
Thus,the expression simplifies to:
$f(11) = 0 + 0 + 0 + 11 - 3 = 8$.

(B) Given the functional equation $f(x+y) = f(x) + f(y) + xy$.
Let $f(x) = ax^2 + bx + c$.
Substituting this into the equation: $a(x+y)^2 + b(x+y) + c = (ax^2 + bx + c) + (ay^2 + by + c) + xy$.
$a(x^2 + 2xy + y^2) + bx + by + c = ax^2 + ay^2 + bx + by + 2c + xy$.
$ax^2 + 2axy + ay^2 + bx + by + c = ax^2 + ay^2 + bx + by + 2c + xy$.
Comparing coefficients of $xy$,we get $2a = 1$,so $a = 1/2$.
Comparing constant terms,$c = 2c$,so $c = 0$.
Given $f(1) = 2$,we have $a(1)^2 + b(1) = 2$,so $1/2 + b = 2$,which gives $b = 3/2$.
Thus,$f(x) = \frac{1}{2}x^2 + \frac{3}{2}x = \frac{x(x+3)}{2}$.
We need to calculate $\sum_{k=1}^{10} f(k) = \sum_{k=1}^{10} (\frac{1}{2}k^2 + \frac{3}{2}k) = \frac{1}{2} \sum_{k=1}^{10} k^2 + \frac{3}{2} \sum_{k=1}^{10} k$.
Using summation formulas: $\sum_{k=1}^{n} k^2 = \frac{n(n+1)(2n+1)}{6}$ and $\sum_{k=1}^{n} k = \frac{n(n+1)}{2}$.
For $n=10$: $\sum k^2 = \frac{10(11)(21)}{6} = 385$ and $\sum k = \frac{10(11)}{2} = 55$.
Sum $= \frac{1}{2}(385) + \frac{3}{2}(55) = 192.5 + 82.5 = 275$.

(C) Given the functional equation $f(x+y)=f(x)+12y$.
Setting $x=0$,we get $f(y)=f(0)+12y$.
Since $f(1)=6$,we have $6=f(0)+12(1)$,which implies $f(0)=-6$.
Thus,$f(x)=12x-6$.
Now,we calculate the sum $\sum_{r=1}^n f(r) = \sum_{r=1}^n (12r-6)$.
$= 12 \sum_{r=1}^n r - \sum_{r=1}^n 6$.
$= 12 \frac{n(n+1)}{2} - 6n$.
$= 6n(n+1) - 6n = 6n^2+6n-6n = 6n^2$.

(C) Given the functional equation $f(x+y)=f(x)+f(y)$ for all $x, y \in R$,this is Cauchy's functional equation,which implies $f(x)=ax$ for some constant $a \in R$.
Given $f(1)=7$,we substitute $x=1$ into $f(x)=ax$ to get $7=a(1)$,so $a=7$.
Thus,the function is $f(x)=7x$.
We need to calculate the sum $\sum_{r=1}^{n} f(r) = \sum_{r=1}^{n} 7r$.
Taking the constant $7$ out,we get $7 \sum_{r=1}^{n} r$.
Using the formula for the sum of the first $n$ natural numbers,$\sum_{r=1}^{n} r = \frac{n(n+1)}{2}$.
Therefore,$\sum_{r=1}^{n} f(r) = 7 \times \frac{n(n+1)}{2} = \frac{7n(n+1)}{2}$.

(A) Given $f(x) = ax^2 + bx + c$ and $a + b + c = 3$,we have $f(1) = a(1)^2 + b(1) + c = a + b + c = 3$.
Given the functional equation $f(x + y) = f(x) + f(y) + xy$.
Putting $y = 1$,we get $f(x + 1) = f(x) + f(1) + x$.
Substituting $f(1) = 3$,we have $f(x + 1) - f(x) = x + 3$.
Summing from $x = 1$ to $n - 1$:
$\sum_{x=1}^{n-1} (f(x+1) - f(x)) = \sum_{x=1}^{n-1} (x + 3)$.
This is a telescoping sum: $f(n) - f(1) = \frac{(n-1)n}{2} + 3(n-1)$.
Since $f(1) = 3$,we have $f(n) = 3 + \frac{n^2 - n}{2} + 3n - 3 = \frac{n^2 + 5n}{2}$.
Now,calculate $\sum_{n=1}^{10} f(n) = \sum_{n=1}^{10} (\frac{n^2}{2} + \frac{5n}{2})$.
Using summation formulas $\sum n^2 = \frac{n(n+1)(2n+1)}{6}$ and $\sum n = \frac{n(n+1)}{2}$:
$\sum_{n=1}^{10} f(n) = \frac{1}{2} \left( \frac{10 \cdot 11 \cdot 21}{6} \right) + \frac{5}{2} \left( \frac{10 \cdot 11}{2} \right)$.
$= \frac{385}{1} + \frac{275}{2} = 192.5 + 137.5 = 330$.

(C) Given the function $f(x) = x^2 + ax + b$.
First,calculate $f(0)$:
$f(0) = (0)^2 + a(0) + b = b$.
Now,substitute this into the given condition $f(f(0)) = 0$:
$f(b) = 0$.
Substitute $x = b$ into the function $f(x)$:
$b^2 + ab + b = 0$.
Since $b \neq 0$,we can divide the entire equation by $b$:
$b + a + 1 = 0$.
Rearranging the terms,we get:
$a + b = -1$.

(B) Given the function $f(xy) = \frac{f(x)}{y}$ for all $x, y > 0$.
We are given $f(30) = 20$.
To find $f(40)$,we can express $40$ as $30 \times \frac{4}{3}$.
Substituting $x = 30$ and $y = \frac{4}{3}$ into the given functional equation:
$f(30 \times \frac{4}{3}) = \frac{f(30)}{\frac{4}{3}}$.
$f(40) = f(30) \times \frac{3}{4}$.
Substituting the value $f(30) = 20$:
$f(40) = 20 \times \frac{3}{4} = 5 \times 3 = 15$.

(D) Given,$f(x+1)+f(x-1)=\sqrt{2} f(x)$ ---$(i)$
Replace $x$ by $x+1$ in $(i)$:
$f(x+2)+f(x)=\sqrt{2} f(x+1)$ ---(ii)
Replace $x$ by $x-1$ in $(i)$:
$f(x)+f(x-2)=\sqrt{2} f(x-1)$ ---(iii)
Adding (ii) and (iii):
$f(x+2)+f(x-2)+2f(x)=\sqrt{2}[f(x+1)+f(x-1)]$
Substitute the value from $(i)$ into the right side:
$f(x+2)+f(x-2)+2f(x)=\sqrt{2}(\sqrt{2} f(x))$
$f(x+2)+f(x-2)+2f(x)=2f(x)$
$f(x+2)+f(x-2)=0$

(C) Given that,$f(x) = \frac{a^x + a^{-x}}{2}, (a > 2) . . . . (i)$
We need to find $f(x + y) + f(x - y)$.
$f(x + y) = \frac{a^{x+y} + a^{-(x+y)}}{2}$
$f(x - y) = \frac{a^{x-y} + a^{-(x-y)}}{2}$
Now,$f(x + y) + f(x - y) = \frac{a^{x+y} + a^{-x-y}}{2} + \frac{a^{x-y} + a^{-x+y}}{2}$
$= \frac{a^x \cdot a^y + a^{-x} \cdot a^{-y} + a^x \cdot a^{-y} + a^{-x} \cdot a^y}{2}$
$= \frac{a^x(a^y + a^{-y}) + a^{-x}(a^y + a^{-y})}{2}$
$= \frac{(a^x + a^{-x})(a^y + a^{-y})}{2}$
$= 2 \cdot \left( \frac{a^x + a^{-x}}{2} \right) \cdot \left( \frac{a^y + a^{-y}}{2} \right)$
$= 2 \cdot f(x) \cdot f(y)$

(A) Given that $f: R \rightarrow R$ satisfies the Cauchy functional equation $f(x+y)=f(x)+f(y)$ for all $x, y \in R$ and $f(1)=5$.
Since $f(x+y)=f(x)+f(y)$,it follows that $f(n)=n \cdot f(1)$ for any integer $n$.
Given $f(1)=5$,we have $f(n)=5n$.
We need to find the sum $\sum_{r=1}^n f(r) = \sum_{r=1}^n 5r$.
This is equal to $5 \sum_{r=1}^n r$.
Using the formula for the sum of the first $n$ natural numbers,$\sum_{r=1}^n r = \frac{n(n+1)}{2}$.
Therefore,$\sum_{r=1}^n f(r) = 5 \times \frac{n(n+1)}{2} = \frac{5n(n+1)}{2}$.

(A) Given the functional relation: $(f(x))^2 = f(x^2) + f(1)$.
We test the options by substituting $f(x) = x + \frac{1}{x}$ into the equation.
For the Right Hand Side $(RHS)$: $f(x^2) + f(1) = (x^2 + \frac{1}{x^2}) + (1 + \frac{1}{1}) = x^2 + \frac{1}{x^2} + 2$.
This can be rewritten as $(x + \frac{1}{x})^2$.
Since $(f(x))^2 = (x + \frac{1}{x})^2$,the Left Hand Side $(LHS)$ equals the $RHS$.
Thus,$f(x) = x + \frac{1}{x}$ is the correct solution.

(D) Given the function $f: N \times N \rightarrow N$ with the conditions:
$f(1,1) = 2$
$f(m+1, n) = f(m, n) + 2(m+n)$
$f(m, n+1) = f(m, n) + 2(m+n-1)$
To find $f(2,2)$,we can use the given recurrence relations:
First,calculate $f(2,1)$ using the first relation with $m=1, n=1$:
$f(2,1) = f(1,1) + 2(1+1) = 2 + 2(2) = 2 + 4 = 6$
Next,calculate $f(2,2)$ using the second relation with $m=2, n=1$:
$f(2,2) = f(2,1) + 2(2+1-1) = 6 + 2(2) = 6 + 4 = 10$
Alternatively,using the other path:
$f(1,2) = f(1,1) + 2(1+1-1) = 2 + 2(1) = 4$
$f(2,2) = f(1,2) + 2(1+2) = 4 + 2(3) = 4 + 6 = 10$
Thus,$f(2,2) = 10$.

(C) Let $f(x) = a_0 x^n + a_1 x^{n-1} + \ldots + a_n$.
Given the functional equation $f(x) \cdot f\left(\frac{1}{x}\right) = f(x) + f\left(\frac{1}{x}\right)$.
This equation is satisfied by $f(x) = 1 \pm x^n$.
Given $f(4) = 257$,we have $1 \pm 4^n = 257$.
This implies $4^n = 256$,so $n = 4$.
Thus,$f(x) = 1 + x^4$.
Finally,$f(3) = 1 + 3^4 = 1 + 81 = 82$.

(C) Given that $f(x + y) = f(x) + f(y)$ for all $x, y \in R$,this is Cauchy's functional equation,which implies $f(x) = cx$ for some constant $c$.
Since $f(1) = 7$,we have $c(1) = 7$,so $c = 7$.
Thus,$f(x) = 7x$.
Now,we are given $\sum_{r=1}^{n} f(r) = 14112$.
Substituting $f(r) = 7r$,we get $\sum_{r=1}^{n} 7r = 14112$.
$7 \sum_{r=1}^{n} r = 14112$.
$7 \cdot \frac{n(n + 1)}{2} = 14112$.
$\frac{n(n + 1)}{2} = \frac{14112}{7} = 2016$.
$n(n + 1) = 4032$.
Since $63 \times 64 = 4032$,we have $n = 63$.

(D) Given: $f(0)=0, f(1)=1, f(2)=2$ and the recurrence relation $f(x)=f(x-2)+f(x-3)$ for $x \ge 3$.
We calculate the values step by step:
$f(3) = f(1) + f(0) = 1 + 0 = 1$
$f(4) = f(2) + f(1) = 2 + 1 = 3$
$f(5) = f(3) + f(2) = 1 + 2 = 3$
$f(6) = f(4) + f(3) = 3 + 1 = 4$
$f(7) = f(5) + f(4) = 3 + 3 = 6$
$f(8) = f(6) + f(5) = 4 + 3 = 7$
$f(9) = f(7) + f(6) = 6 + 4 = 10$
Therefore,$f(9) = 10$.

(B) The given conditions are $f(x+y) = f(x) + f(y)$ (Cauchy's functional equation) and $f(xy) = f(x)f(y)$.
For $x, y \in \mathbb{Q}$,the only solutions to these equations are the identity function $f(x) = x$ and the zero function $f(x) = 0$.
$1$. If $f(1) = 0$,then $f(x) = f(x \cdot 1) = f(x)f(1) = 0$ for all $x$.
$2$. If $f(1) \neq 0$,then $f(1) = f(1 \cdot 1) = f(1)^2$,which implies $f(1) = 1$. By induction,$f(n) = n$ for all $n \in \mathbb{N}$,and consequently $f(x) = x$ for all $x \in \mathbb{Q}$.
Thus,there are exactly $2$ such functions.
Hence,option $B$ is correct.

(D) The given equation $f(x+y) = f(x) + f(y)$ is Cauchy's functional equation defined over the set of integers $\mathbb{Z}$.
For any $x \in \mathbb{Z}$,we can write $f(nx) = nf(x)$ for any $n \in \mathbb{Z}$.
Specifically,for $x = 1$,we have $f(n) = f(n \cdot 1) = n \cdot f(1)$.
Let $f(1) = c$,where $c$ is any integer constant because the codomain is $\mathbb{Z}$.
Thus,$f(n) = cn$ for any $c \in \mathbb{Z}$.
Since there are infinitely many choices for the integer $c$,there are infinitely many such functions.

(D) Given the functional equation $f(x+y)=f(x) \cdot f(y)$.
Setting $x=4$ and $y=0$,we get $f(4+0)=f(4) \cdot f(0)$,which implies $f(4)=f(4) \cdot f(0)$. Since $f(x)$ is differentiable,$f(x)$ is not identically zero,so $f(0)=1$.
By the definition of the derivative,$f^{\prime}(x) = \lim_{h \rightarrow 0} \frac{f(x+h)-f(x)}{h}$.
Substituting $f(x+h)=f(x) \cdot f(h)$,we have $f^{\prime}(x) = \lim_{h \rightarrow 0} \frac{f(x)f(h)-f(x)}{h} = f(x) \lim_{h \rightarrow 0} \frac{f(h)-1}{h}$.
Since $f(0)=1$,this is $f^{\prime}(x) = f(x) \lim_{h \rightarrow 0} \frac{f(h)-f(0)}{h} = f(x) \cdot f^{\prime}(0)$.
Given $f^{\prime}(4)=24$ and $f^{\prime}(0)=3$,we substitute these into the equation $f^{\prime}(4) = f(4) \cdot f^{\prime}(0)$.
$24 = f(4) \cdot 3$.
Therefore,$f(4) = \frac{24}{3} = 8$.

(B) Given the functional equation $f(x+y)=f(x) \cdot f(y)$ for all $x, y \in R$.
Setting $x=0$ and $y=0$,we get $f(0)=f(0)^2$,which implies $f(0)(f(0)-1)=0$.
Since $f(x) \neq 0$ for all $x$,we must have $f(0)=1$.
By the definition of the derivative at $x=0$,$f^{\prime}(0) = \lim_{h \rightarrow 0} \frac{f(0+h)-f(0)}{h} = \lim_{h \rightarrow 0} \frac{f(h)-1}{h} = 4$.
Now,for any $x$,the derivative $f^{\prime}(x)$ is given by:
$f^{\prime}(x) = \lim_{h \rightarrow 0} \frac{f(x+h)-f(x)}{h} = \lim_{h \rightarrow 0} \frac{f(x)f(h)-f(x)}{h} = f(x) \lim_{h \rightarrow 0} \frac{f(h)-1}{h}$.
Substituting the value of the limit,we get $f^{\prime}(x) = f(x) \cdot 4 = 4f(x)$.
Therefore,$f^{\prime}(6) = 4f(6) = 4 \times 3 = 12$.

(D) Given the functional equation: $f(xy+1)=f(x)f(y)-f(y)-x+2$.
Substitute $y=0$ into the equation:
$f(x(0)+1) = f(x)f(0) - f(0) - x + 2$
Since $f(0)=1$,we have:
$f(1) = f(x)(1) - 1 - x + 2$
$f(1) = f(x) - x + 1$
Rearranging the terms to express $f(x)$:
$f(x) = x + f(1) - 1$
Now,differentiate both sides with respect to $x$:
$\frac{df}{dx} = \frac{d}{dx}(x + f(1) - 1)$
Since $f(1)$ is a constant,its derivative is $0$:
$\frac{df}{dx} = 1 + 0 - 0 = 1$
Therefore,the value of $\frac{df}{dx}$ at $x=e$ is $1$.

(C) Given $f(x) f\left(\frac{1}{x}\right) = f(x) + f\left(\frac{1}{x}\right)$.
Let $f(x) = ax^2 + bx + c$. The given equation is equivalent to $f(x) - 1 = \frac{1}{f(1/x) - 1}$ or similar forms. $A$ standard solution for this functional equation is $f(x) = x^n + 1$ or $f(x) = -x^n + 1$.
Since $f(x)$ is a quadratic function,we take $f(x) = x^2 + 1$.
Calculating the values:
$f\left(\frac{2}{3}\right) = \left(\frac{2}{3}\right)^2 + 1 = \frac{4}{9} + 1 = \frac{13}{9}$.
$f\left(\frac{3}{2}\right) = \left(\frac{3}{2}\right)^2 + 1 = \frac{9}{4} + 1 = \frac{13}{4}$.
Now,$\sqrt{f\left(\frac{2}{3}\right) + f\left(\frac{3}{2}\right)} = \sqrt{\frac{13}{9} + \frac{13}{4}} = \sqrt{\frac{52 + 117}{36}} = \sqrt{\frac{169}{36}} = \frac{13}{6}$.

(C) Given,$f(x)=x^2-2x+4$.
We need to solve $f(x-1)=f(x+1)$.
Substituting $(x-1)$ and $(x+1)$ into the function:
$(x-1)^2-2(x-1)+4 = (x+1)^2-2(x+1)+4$.
Expanding both sides:
$(x^2-2x+1) - 2x + 2 + 4 = (x^2+2x+1) - 2x - 2 + 4$.
Simplifying:
$x^2-4x+7 = x^2-1+4$.
$x^2-4x+7 = x^2+3$.
Subtracting $x^2$ from both sides:
$-4x+7 = 3$.
$-4x = 3-7$.
$-4x = -4$.
$x = 1$.
Thus,the set of values is $\{1\}$.